Problem: Simplify the following expression: $y = \dfrac{-9x^2+41x- 20}{x - 4}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-9)}{(-20)} &=& 180 \\ {a} + {b} &=& &=& {41} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $180$ and add them together. The factors that add up to ${41}$ will be your ${a}$ and ${b}$ When ${a}$ is ${5}$ and ${b}$ is ${36}$ $ \begin{eqnarray} {ab} &=& ({5})({36}) &=& 180 \\ {a} + {b} &=& {5} + {36} &=& 41 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-9}x^2 +{5}x) + ({36}x {-20}) $ Factor out the common factors: $ x(-9x + 5) - 4(-9x + 5)$ Now factor out $(-9x + 5)$ $ (-9x + 5)(x - 4)$ The original expression can therefore be written: $ \dfrac{(-9x + 5)(x - 4)}{x - 4}$ We are dividing by $x - 4$ , so $x - 4 \neq 0$ Therefore, $x \neq 4$ This leaves us with $-9x + 5; x \neq 4$.